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SOLUTIONS TOF.F. CHENSPLASMAPHYSICS
(VERSION 0.5 )
SJTU
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CONTENTS
Preface v
Acknowledgments vii
1 Introduction 1
1.1 1-3 21.2 1-8 31.3 1-9 31.4 1-10 41.5 1-11 4
2 Motion of Particle 5
2.1 2-2 52.2 2-3 62.3 2-7 62.4 2-10 62.5 2-11 62.6 2-15 72.7 2-16 8
iii
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iv CONTENTS yetao1991.wordpress.com
2.8 2-17 82.9 2-18 82.10 2-19 92.11 2-20 102.12 2-21 12
3 Wave in Plasma 13
3.1 4-5 133.2 4-9 143.3 4-16 143.4 4-18 153.5 4-23 153.6 4-26 163.7 4-27 163.8 4-37 163.9 4-38 (Need to be confirmed) 183.10 4-39 193.11 4-40 193.12 4-41 193.13 4-42 193.14 4-43 193.15 4-45 193.16 4-49 193.17 4-50 19
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PREFACE
Last revised: 11/24/2013 If you have any suggestions or ideas on this book, pleasetell me.My E-mail: [email protected] Personal Homepage: http://yetao1991.wordpress.com
ME
Shanghai,China
November, 2013
v
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ACKNOWLEDGMENTS
Thanks All.
T.Y.
vii
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CHAPTER 1
INTRODUCTION
1
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2 INTRODUCTION yetao1991.wordpress.com
Problem 1-3
-2
-1
0
1
2
3
4
5
6 8 10 12 14 16 18 20 22 24
log(k
T)
log(n)
Problem 1-3
(1)
(2.1)
(2.2)
(3)
(4)
(5)(6)
(7)
1
100
1e4
1e6
1e8
1e-1 1e-2 1e-3 1e-4 1e-5 1e-7
Debye length is constant at the line of Nd is constant at the line of
Figure 1.1 Label (1)(2.1)(2.2)(3)(4)(5)(6)(7) correspond to plasma in different condition asthe problem describes.
According to the definition of the Debye Length
D = (0kTene2
)1=2 (1.1)
) log(D) = 12log(
0e2
) +1
2log(kTe) 1
2log(n) (1.2)
) log(kT ) = log(n) + 2 log( D7430
) (kT in eV) (1.3)
Then we can draw the solid straight line in the Figure1.1 with the Debye lengthas parameter ranged from 101 to 107. Points on a certain solid line, named withequi-Debye-length line(analog to equipotential lines in electrostatic) , share a same
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[email protected] 1-8 3
Debye length. Similarily, with the given equation:
ND =4
33Dn (1.4)
=4
3(7430 kT
n)32 (kT in eV) (1.5)
) ND2:7 106 =
(kT )32
n12
(1.6)
) log( ND2:7 106 ) =
3
2log(kT ) 1
2log(n) (1.7)
) log(kT ) = 13log(n) +
2
3log(
ND2:7 106 ) (1.8)
The dot lines in the Fig 1.1 can be named with equi-ND lines, which means theyshare a same value of ND in a dot lines.
As for the usage of this figure, take the point (4) for example, the point(4) fallsin the region enclosed by two solid lines and two dots ones.The two solid lines re-spectively have the Debye length of 103m and 104m. And the dots lines havethe particle numbers ND of 104 and 106. That is tantamount to the fact that (4) hasDebye length 104m < D < 103m and number of particle 104 < ND < 106.
Recall for the criteria for plasmas, ND o 1 is automatically meet since thesmallest number of particle is larger than 100.
Problem 1-8
The Debye length is
D = 69(T
n)12 (T in the unit of K)
= 69 (5 107
1033)12
= 1:54 1011mNaturally, the number of particles contained in a Debye Sphere is :
ND =4
33D n
=4
3 (1:54)3
15
Problem 1-9
Since protons and antiprotons have the same inertia, both of them are fixed. Assumethat protons and antiprotons follows the Maxwellian distribution.
f(u) = Ae(12mu
2+q)=kTe ;
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4 INTRODUCTION yetao1991.wordpress.com
where q equals to e for protons while q equals to e for antiprotons. Moreover,
np(!1) = n1np(!1) = n1:
Then we obtain (np = n1exp(ekT )np = n1exp( ekT )
The Poissons Equation is
"0r2 = "0 @2
@x2= e(np np)
With e=kT 1,"0@2
@x2=
2n1e2kT
) D =r
"0kT
2n1e2= 0:4879m
Problem 1-10
Problem 1-11
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CHAPTER 2
MOTION OF PARTICLE
Problem 2-2
Since A=2, for deuterium ion,
m = 2mp = 3:34 1027kgq = jej = 1:60 1019Coulomb:
Assume that energy can be entirely converted to kinetic energy, then the momentumcan be derived
E = Ek =p2
2m) p =
p2mEk = 1:46 1020kg m=s
The Larmour radius
rL =mv?jqjB =
1:46 10205 1:60 1019 = 0:018m a = 0:6m
So the Larmour radius satisfies the confined-ion condition.
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6 MOTION OF PARTICLE yetao1991.wordpress.com
Problem 2-3
To keep a equibrillium in the y^ direction, the electric force shouldconteract the Lorentz force
Eq = qvB
) E = vB = 106V=m
Problem 2-7
Apply the Gauss Law to obtain the magnitude of electric field at r=a
E l 2a = nel a2="0) E = enea
2"= 9:04 103V=m
vE =E
B= 4:52 103m=s
(Direction:See in the figure)
Problem 2-10
The mass of a deutron ismd = 1875MeV=c
2
The kinetic energy
Ek =1
2mdv
2 ) v = 1:386 106m=s
v? = v cos 45 = 9:8 105m=s
rLmdv?qB
= 0:03m
Problem 2-11
Rm = 4) 1sin2 m
= 4) sin m = 12
) m =
6
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[email protected] 2-15 7
Since the velocity is isotropic distribution, the directionof velocity should distribute uniformly
d = sin dd'
The total solid angle for a sphere is
total = 4
The solid angle for loss cone is
loss = 2
Z 6
0
sin d
Z 20
d' = (1p3
2)4
) the fraction of the trapped isp32 .
Problem 2-15
Define the displacement in polarization direction is xp.So the work done by the electric field is
W = q ~E ~xPThe energy gain rate is
dW
dt= qE
dxpdt
= qEvp:
The change of kinetic energy
dEkdt
=d
dt(1
2mv2E) = mvE
dvEdt
Without loss of generality, let the vE in the same direction of E
vE = EB
) dvEdt
= 1B
dE
dt
) dEkdt
= mvE( 1B
dE
dt) = qEvp
vp =m
qB
1
B
dE
dt
Replace 1!c = mqB Thus,vp = 1
!cB
dE
dt
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8 MOTION OF PARTICLE yetao1991.wordpress.com
Problem 2-16
a) The Larmor frequency of electron is
!e =eB
m=
1:6 1019 19:109 1031 = 1:76 10
11rad=sec
b) the Larmor frequency of ion is
!i =eB
mi=
1:6 1019 11:67 1027 = 9:58 10
7rad=sec
Since !0 = 109rad=sec, !e !0 !i. The motion of electron is adiabatic,while that of ion not.
Problem 2-17
Since = mv2?
2B is conservative under this condition, it is easy to derive:
mv2?2B
=mv02?2B0
where 12mv2? = 1keV; B = 0:1T;B0 = 1T , so
1
2mv02? = 10keV:
When collision happens, the direction of motion distorts, so v? = vk. Then thekinetic energy is
1
2mv002? = 5keV
Implement the adiabatic characteristic of , we know that
mv002?2B0
=mv0002?2B0
) 12mv0002? = 0:5keV
Finally, the energy is
E = E? + Ek =1
2mv0002? +
1
2mv002k = 5:5keV
Problem 2-18
a) At a certain moment, we calculate the motion in one periodic circular motion tocertify the invariance of . In this period, we assume that the Larmour radius doesnot change with minor deviation of magnetic field-B. So
s = r2L = m2v2?q2B2
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[email protected] 2-19 9
d
dt =
dB
dt S = m
2v2?q2B2
dB
dt= "(induced potential)
So the change of the energy of the particle in one period is
W = q" = m2v2?qB2
dB
dt
Within this period of gyration, the change of magnetic field is
B =dB
dt =
dB
dt
2m
qB
So after a period, the energy is
E0k =1
2mv2? + W = (1 +
2m
qB2dB
dt)1
2mv2?
B0 = B +B = (1 +2m
qB2dB
dt)B
0 =E0kB0
=E0B0
=
So is invariant for both ion and electron.b) Assume that 12mv
2? = KT;
12mv
2k = KT at initial moment. Considering
=12mv
2?
B=
12mv
02?
B0;
and B = 13B0, so 12mv
02? = 3KT . Furthermore, we assume that vk remains un-
changed,1
2mv2k = KT
kT? = 3Kev; kTk = 1KeV
Problem 2-19
The magnetic field is B = 0I2r , so the gradient is :
rB = @B@r
=0I
2r2
Apply the equipartition theorem of Maxwellian gas, the total energy is
" =1
2m(p2x + p
2y + p
2z) =
3
2kT
) "? =1
2m(p2x + p
2y) = kT
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10 MOTION OF PARTICLE yetao1991.wordpress.com
The perpendicular velocity is :
) v? =r
2kT
m
) rL =mv?eB
) vrB =1
2v?rL
rBB
=kT
eBr;Upwards
* vk =rkT
m) vR =
mv2keBr
=kT
eBr;Upwards
) vR+rB =2kT
eBr
Area of the top surface of the cyclinder is :
s = [(R+a
2)2 (R a
2)2] = 2Ra
The number of charged particle, which hit the surface s with in time of dt is:
N = n v s dt:
So the hit rate isdN
dt= n v s:
Since R a, the drift velocity with [R a2 ; R + a2 ] region can be considered asuniform. So the accumulation rate is
Racc = n vR+rB s e = 4kTnaB
= 20Coulomb=S
Problem 2-20
a) v2k + v2? is invariant because of the conservation of energy. At z = 0, Bz = B0,
v2? =23v
2,v2k =13v
2,
=mv2?2B0
=mv2
2B0:
When the electron reflects, vk = 0, then v2? = v2
=mv2
3B0) Bz = 3
2B0
) 1 + 2z2 ) z = p2
2
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[email protected] 2-20 11
b)
* = mv2?
2Bz) B0(1 + 2z2) = 1
2mv2?
) v2k = v2 v2? =0B
m 20B
m2z2
) (dzdt
)2 =0B
m(1 22z2)
dz
dt=
r0B0m
p1 22z2
) z =p2
2sin(
r20B0m
t+ )
And this equation can describe the trajectory of the particle.
c) It is apparent that the gyration frequency isq
20B0m from the the equation of
motion.d) Claim: =
q20B0m t+ , z =
p2
2 sin
J =
Z ba
vkdz
where a = p2
2 ; b =p2
2 . Thus,
a = 2; b =
2
J =
Z ba
vkdz =Z
vkvkdt =1
rm
2B0
Z ba
v2kd
where
dt =1
rm
2B0d;
v2k = (dz
dt)2 =
B0m
sin2 :
) J = 1
rm
2B0
Z 2
2
B0m
sin2 d
=
2
rB02m
= constant
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12 MOTION OF PARTICLE yetao1991.wordpress.com
Problem 2-21
a) According to the Ampere theorem, the magnetic field can be obtained
2r B = 0I ) B = 0I2r
In cylinder coordinate:
rB = @B@r
=0I
2r2
) vrB =1
2v?rL
rBB
;
where rL = mv?eB ; v?0 = v?.
) vrB =1
2v?0
mv?0eB
1
r=
mv2?00Ie
Besides,
vR =mv2ke
1
RcB
=mv2ke
1
Rc 0I2RC=
2mv2k0Ie
) ~v = ~vR + ~vrB =3mv2k00Ie
(z^)
(vk0 = v?0)
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CHAPTER 3
WAVE IN PLASMA
Problem 4-5
According to the dispersion relation
!2 = !2p +3KTem
k2
Now we calculate the unknown value in the equation
!p 2pn = 5:62 109rad=sec
! = 2f = 6:908 109rad=sec3kTem
=3 100 1:6E19
9:109 1031 = 5:27 1013
) k2 =!2 !2p
3kTem
= 3:05 105 ) k = 552:377
) = 2k
= 1:14 102m
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14 WAVE IN PLASMA yetao1991.wordpress.com
Problem 4-9
The critical density is
n0 =m"0!
2
e2= 1:12 1015m3
Problem 4-16
If the motion of ion is neglected, the dispersion relation of electron is
c2k2
!2= 1 !
2p
!2!2 !2p!2 !2h
1) The resonance of X-wave is found by setting k ! 1. So the dispersion relationcan be rewrite into
c2k2 = !2 !2p!2 !2p!2 !2h
Differentiate both sides of the equation
2kc2dk = 2!d! 2!(!2p !2h
!2 !2h!2pd!
So the group velocity is
vg =dw
dk=
kc2
![1 +!2c!
2p
(!2!2h)2]
When k !1 ,which implies that ! = !h.So 1 +
!2c!2p
(!2!2h)2! 1; k ! 1 and 1 + !
2c!
2p
(!2!2h)2has a higher order than k. Thus,
vg = 0 at resonance point.2) The cut-off X-wave is found by setting k=0, then
1 !2p
!2!2 !2p!2 !2h
) ! = !R or !L
At this point, !(1 +!2c!
2p
(!2!2h)2) is a finite value. As a result
vg =kc2
![1 +!2c!
2p
(!2!2h)2]jk=0 = 0
) Q:E:D:
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[email protected] 4-18 15
Problem 4-18
For L-wave,c2k2
!2= 1 !
2p=!
2
1 + !c=!
And
!2p =n0e
2
m"0; !c =
eB
m
The cut-off is found when k = 0. With the provided condtion f = 2:8GHz;B0 =0:3T , the critical density is
n0 =m"0e2
[(2f)2 + 2feB
m]
= 3:89 1017m3
Problem 4-23
R-wave
k2 =!2
c2 !
2p
c2(1 !c=!p)L-wave
k2 =!2
c2 !
2p
c2(1 + !c=!p)
Thus
kR =!
c
s1 !
2p=!
2
1 !c! !
c(1 1
2
!2p=!2
1 !c!) !
c[1 1
2
!2p!2
(1 +!c!)]
kL =!
c
s1 !
2p=!
2
1 + !c! !
c(1 1
2
!2p=!2
1 + !c!) !
c[1 1
2
!2p!2
(1 !c!)]
The difference of the phase is twice of the Faraday rotation angle
=1
2
180
(kL kr)
=180
2
1
c(
1
42)e2
m"0
e
mB(Z)n(z)
=90
1
c(
1
42)(0c)2
e3
m2"0B(z)n(z)
=90
e3
42m2"0c3B(z)n(z)20(degree)
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16 WAVE IN PLASMA yetao1991.wordpress.com
And90
e3
42m2"0c3= 1:5 1011degree = 2:62 1013rad
In conclusion
=180
2
Z L0
(kL kR)dz = 1:5 10110Z L0
B(z)n(z)dz
(in the unit of degree)
Problem 4-26
a)
va =Bp0
= 2:18 108m=s
b) The Alfven wave represents for phase velocity. And phase velocity did not carryinformation. So it does not mean that wave can travel faster than light.
Problem 4-27
= n0M = 1:67 1019
vA =Bp0
= 2:18 104m=s
Problem 4-37
a) Consider the elastical collision from ion, the equation of eletrons motion
m@ve@t
= eE mve
Linearize the equation
im!ve = eE mve ) j1 = n0eve = n0e2E1
im(! + i)
The equation of the transverse wave is
(!2 c2k2)E1 = i!j1="Insert j1 into the wave equation, we get
(!2 c2k2)E1 = !! + i
!2pE1
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[email protected] 4-37 17
) !2 c2k2 = !!2p
! + i
) c2k2
!2= 1 !
2p
!(! + i)
) Q:E:D:
b) Apply the previous result
c2k2 = !2 !2p
1 + i !
When ! or ! 1,then
(1 + i
!)1 1 i
!
So the dispersion relation turns to
c2k2 = !2 !2p + i!2p!
Assume that ! = a+ bi, then
!2p + c2k2 = (a2 b2) + 2abi+ i !
2p
a+ bi
= (a2 b2) + !2pvb
a2 + b2+ i(2ab+
!2pa
a2 + b2)
The image part should be zero, that is
ib =!2pv
2(a2 + b2)i = Im(!)i
So the damping rate:
= Im(!) = !2p
2j!j2
c) With previous conclusion of a), we get
k2 =!2 !2p
c2+ i
!2p
c2!
Let k = e+ di, then k2 = e2 d2 + 2edi, we get8
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18 WAVE IN PLASMA yetao1991.wordpress.com
Problem 4-38 (Need to be confirmed)
The loss part of energy will heat up the electron to oscillation. So we firstly need toderive the vibration motion due to the microwave. The wave-length of the microwaveis = 0:3m So the frequency is
! =2c
= 6:28 109rad=sec
The collision frequency is
= nn = 102=sec
So the equation of motion of electron is
m@ve@t
= eE mve
Linearize the equation to be
i!mve = eE mve ) ve = eEim(! + i)
Then the dispersion raltion is
k2 =!2 !2p
c2+!2p
!2ci
k = Re(k) + Im(k)i
And microwave in plasma is
E = E0ei(kr!t) = E0ei(Re(k)r!t)eIm(k)r
The term eIm(k)r means the wave decay when penetrate the ionosphere. The ratioof outgoing energy v.s. incident energy is
A =E2outE2ini
= (eIm(k)R1
eIm(k)R2)2 = e2Im(k)(R1R2)
And R1 R2 = 100km = 105m
!2p =nee
2
"0m= 3:17 1014
Im(k) =
2c
!2p!2
1q1 !2p!2
= 1:34 1012
) A = e2:69107 = 0:999999732So the loss fraction is
1A 0
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[email protected] 4-39 19
Problem 4-39
Unchecked.
Problem 4-40
Problem 4-41
Problem 4-42
Problem 4-43
Problem 4-45
Problem 4-49
Problem 4-50
Unchecked.